# 设计一个找到数据流中第 k 大元素的类（class）。注意是排序后的第 k 大元素，不是第 k 个不同的元素。
#  请实现 KthLargest 类：
#  KthLargest(int k, int[] nums) 使用整数 k 和整数流 nums 初始化对象。
#  int add(int val) 将 val 插入数据流 nums 后，返回当前数据流中第 k 大的元素。
#
#  示例：
# 输入：
# ["KthLargest", "add", "add", "add", "add", "add"]
# [[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
# 输出：
# [null, 4, 5, 5, 8, 8]
#
# 解释：
# KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
# kthLargest.add(3);   // return 4
# kthLargest.add(5);   // return 5
# kthLargest.add(10);  // return 5
# kthLargest.add(9);   // return 8
# kthLargest.add(4);   // return 8
import heapq
from typing import List


class KthLargest:

    def __init__(self, k: int, nums: List[int]):
        self.k = k
        self.minHeap = nums[:min(len(nums), k)]
        heapq.heapify(self.minHeap)
        if len(nums) > k:
            for n in nums[k:]:
                if n > self.minHeap[0]:
                    heapq.heapreplace(self.minHeap, n)

    def add(self, val: int) -> int:
        if len(self.minHeap) >= self.k:
            if val > self.minHeap[0]:
                heapq.heapreplace(self.minHeap, val)
        else:
            heapq.heappush(self.minHeap, val)
        return self.minHeap[0]


if __name__ == "__main__":
    # kthLargest = KthLargest(3, [4, 5, 8, 2])
    # print(kthLargest.add(3))   # return 4
    # print(kthLargest.add(5))   # return 5
    # print(kthLargest.add(10))  # return 5
    # print(kthLargest.add(9))   # return 8
    # print(kthLargest.add(4))   # return 8

    # [[1, []], [-3], [-2], [-4], [0], [4]]
    kthLargest = KthLargest(1, [])
    print(kthLargest.add(-3))   # return -3
    print(kthLargest.add(-2))   # return -2
    print(kthLargest.add(-4))  # return -2
    print(kthLargest.add(0))   # return 0
    print(kthLargest.add(4))   # return 4
